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conversion mg/Nm3 to PPM 1
- Thread starter yulinios
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SO2 has a density of 2.9268 mg/mL at NTP (0 deg C, 1 atm). Thus, 1200 mg would have a volume of 1200/2.9268 = 410 mL.
Since the gases are considered ideal, the given operating conditions wouldn't affect this result.![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
410 mL/m[sup]3[/sup] = 410 ppm (v/v)
Since the gases are considered ideal, the given operating conditions wouldn't affect this result.
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yulinios:
One thing to clarify here. The analyzer indicates +/- 1,200 mg/Nm3. This is at standard conditions and the temperature and pressure conditions cited are merely the characterization of the environment the analyzer is sampling. I believe this is part of what 25362 means by his last statement, but I thought it could bear repeating and a little clarification. The theoretical equation is
(mg/m3)*(1/(1000*MW))*(0.0224m3/mol)= m3 of gas A/m3 of gas mixture
Where mg/m3 is the measured concentration from the analyzer, MW is the molecular weight of the chemical in question in grams/mol and 0.0224 is the volume of a single mole of an ideal gas at standard conditions (i.e., 0C and 1 atm.). The result of this calculation is cubic meters of the particular chemical species of interest per cubic meter of mixed gas. To convert to ppm, multiply the result by 10^6. Thus the final equation is:
(mg/m3)*(22.4/MW)= PPM or 1,200*(22.4/64.06) = 420 ppm
It appears the difference here between my result and 25362's is that I calculate a pure vapor phase density of 2.860 mg/ml, not the quantity used by 25362. I would expect this is a tabulated value based on laboratory results and is representative of the error introduced by the ideal gas assumption.
One thing to clarify here. The analyzer indicates +/- 1,200 mg/Nm3. This is at standard conditions and the temperature and pressure conditions cited are merely the characterization of the environment the analyzer is sampling. I believe this is part of what 25362 means by his last statement, but I thought it could bear repeating and a little clarification. The theoretical equation is
(mg/m3)*(1/(1000*MW))*(0.0224m3/mol)= m3 of gas A/m3 of gas mixture
Where mg/m3 is the measured concentration from the analyzer, MW is the molecular weight of the chemical in question in grams/mol and 0.0224 is the volume of a single mole of an ideal gas at standard conditions (i.e., 0C and 1 atm.). The result of this calculation is cubic meters of the particular chemical species of interest per cubic meter of mixed gas. To convert to ppm, multiply the result by 10^6. Thus the final equation is:
(mg/m3)*(22.4/MW)= PPM or 1,200*(22.4/64.06) = 420 ppm
It appears the difference here between my result and 25362's is that I calculate a pure vapor phase density of 2.860 mg/ml, not the quantity used by 25362. I would expect this is a tabulated value based on laboratory results and is representative of the error introduced by the ideal gas assumption.
To Fizzhead, you guessed right. The density was taken from Perry VI Table 3-30. I prefer to use tabulated NTP or STP data, whenever available, as in the case of SO2, rather than using the ideal gas law's 22.4 m[sup]3[/sup]/mol.
In this manner errors, especially for gaseous molecules composed of several atoms, are minimized.
In this manner errors, especially for gaseous molecules composed of several atoms, are minimized.
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